





Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
2(x) are any two (linearly independent) solutions of a linear, homogeneous second order differential equation then the general solution y cf(x), is y cf(x) = Ay 1(x)+By 2(x) where A, B are constants. We see that the second order linear ordinary differential equation has two arbitrary constants in its general solution.
Typology: Lecture notes
1 / 9
This page cannot be seen from the preview
Don't miss anything!






2.1 Introduction: Many practical problems in engineering give rise to second order differential equations of the form, (2.1) Where are constant coefficients and is a given function of. Case I, Let , Equation (2.1) becomes (2.2) Let be two solutions of the equation and Adding these two lines together, we get Now
Therefore, the equation can be written as Which is our original solution with replaced by. therefore if are solutions of the equation (2.2) , so also is Now if equation (2.2) we get the first order equation Solving this by the method of separating the variables, we have If In the same way, will be a solution of the second order equation (2.2) if it satisfies this equation. Now , if And substituting these expressions for the differential coefficients in the left-hand side of the equation, we get
Example 2 Solve the differential equation The Auxiliary equation is . The solution is (2.3) Real and Equal Roots Let ua take the differential equation The Auxiliary equation is . and their two terms would combine to give. But every second order differential equation has two arbitrary constants. In fact it can be shown that also satisfies the equation. So that the complete general solution is of the form In general, if the auxiliary equation has real and equal roots, giving , the solution of the differential equation is Example 1. Solve Solution
The Auxiliary equation is . (2.4) Complex Roots to the Auxiliary Equation Suppose the roots of the auxiliary equation are complex say Then the solution Now from our previous work on complex numbers we know that Our solution above can therefore be written Where Therefore , if , the solution can be written in the form
(This is like when and ) (i) (ii) But from hyperbolic functions Adding (i) and (ii) Similarly , subtracting Example 1 Solve Solution Auxiliary equation : Example 2 Solve
Solution Auxiliary equation: Case II, In the equation Where , is the extra function yet to be found is called the complementary function (C.F) is called the particular integral (P.I) Note: The complete general solution is given by (a) The complementary function is obtained by solving the equation with , as in the previous part of the programme. This will give one of the following types of solution (b) The particular integral (P.I) is found by assuming the general form of the function on the right-hand side of the given equation, substitute this in the